LeetCode-Sort Characters By Frequency

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本文介绍了一种基于字符出现频率的字符串排序算法，通过桶排序和哈希表实现，确保相同频率的字符保持原有顺序，适用于多种编程场景。

Description：
Given a string, sort it in decreasing order based on the frequency of characters.

Example 1:

Input:
"tree"

Output:
"eert"

Explanation:
'e' appears twice while 'r' and 't' both appear once.
So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.

Example 2:

Input:
"cccaaa"

Output:
"cccaaa"

Explanation:
Both 'c' and 'a' appear three times, so "aaaccc" is also a valid answer.
Note that "cacaca" is incorrect, as the same characters must be together.

Example 3:

Input:
"Aabb"

Output:
"bbAa"

Explanation:
"bbaA" is also a valid answer, but "Aabb" is incorrect.
Note that 'A' and 'a' are treated as two different characters.

题意：按照字符出现的次序，对字符串进行排序，即字符在字符串中出现次数最多的，排序后处在首位；

解法：首先，利用桶排序将出现次数相同的字符放在同一个List中；构建哈希表<key=字符出现的次数，value=List（存储的是出现次数相同的字符）>；从字符串的长度开始遍历，依次将出现次数最多的字符取出（长度即为key的大小），存储在返回结果中；

Java

class Solution {
    public String frequencySort(String s) {
        int[] count = new int[256];
        for (char ch : s.toCharArray()) {
            count[ch]++;
        }

        Map<Integer, List<Character>> map = new HashMap<>();
        for (int i = 0; i < count.length; i++) {
            if (count[i] == 0) {
                continue;
            }
            int cnt = count[i];
            if (!map.containsKey(cnt)) {
                map.put(cnt, new ArrayList<Character>());
            }
            map.get(cnt).add((char)i);
        }

        StringBuilder sb = new StringBuilder();
        for (int i = s.length(); i > 0; i--) {
            if (!map.containsKey(i)) {
                continue;
            }
            List<Character> list = map.get(i);
            for (Character ch : list) {
                for (int len = 0; len < i; len++) {
                    sb.append(ch);
                }
            }
        }

        return sb.toString();
    }
}